Question

Using the Part 1 of Fundamental Theorem of Calculus, explain how can we find the derivative of the given function;
$y = \int {\sqrt {5t + \sqrt t } dt}$; from the values $2$to $\tan (x)$

Answer 1

When we get a function as one of the limits, we may need to simplify the problem by using some methods of substitution. Substitution may lead to using chain rule. Chain rule helps us simplify the process of derivation involving more than a single variable. Remember than when we substitute, after finding the answer do not forget to bring back the original variable.

Complete step by step answer:
Write down the equation that is given in our question;
$y = \int\limits_2^{\tan (x)} {\sqrt {5t + \sqrt t } dt}$
The first step we can try is to use chain substitution. So by chain rule, let our functions be defined as;
$F(x) = \int\limits_2^x {\sqrt {5t + \sqrt t } dt}$
Then comparing with the question we define a chain substitution;
$h(x) = \tan (x)$
Now by the Fundamental Theorem of Calculus, we can directly write;
$F(x) = \int\limits_2^x {\sqrt {5t + \sqrt t } dt}$ $\Rightarrow F'(x) = \sqrt {5x + \sqrt x }$
Next we have to take the derivation of $h(x)$;
$h'(x) \Rightarrow h'(x) = {\sec ^2}(x)$
Now if we go back to the question we can write;
$\int\limits_2^{\tan (x)} {\sqrt {5t + \sqrt t } dt} = F(h(x))$
By taking the Chain rule now, we can write the previous equation as;
$\dfrac{d}{{dx}}(\int\limits_2^{\tan (x)} {\sqrt {5t + \sqrt t } dt} ) = F'(h(x)) \times h'(x)$
So the final answer will be;
$\therefore \dfrac{{dy}}{{dx}} = \sqrt {5\tan (x) + \sqrt {\tan (x)} } \times {\sec ^2}(x)$

Note: Chain rules are applied in two different scenarios both involving two functions for example; $f(x),\;h(x)$
Assume that both the given functions are differentiable.
(i) Scenario 1:
If the function is defined as; $F(x) = (f \circ h)(x)$
It implies that the derivative of $F(x)$will be as follows:
$F'(x) = f'(h(x)) \times h'(x)$
(ii) Scenario 2:
When one variable is substituted multiple times as in the inner term is again assigned as a function;
$y = f(u)$, $u = g(x)$
This means that to find derivative of $y$, we can use the following formula:
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$